Write down the equation for the capacitance of a parallel plate capacitor with d

Question

Write down the equation for the capacitance of a parallel plate capacitor with dielectric material inside. Identify the meaning of each symbol m the equation. A parallel plate capacitor has plates with dimensions 1.00 cm times 40.0 cm. The separation between plates is 0.200 m. If the capacitor filled with Mica with dielectric constant of 7.00, calculate the capacitance. If the plates were connected to 12.0 V battery, calculate the magnitude of the charge on a plate. Calculate the total energy stored in the capacitor when connected to 12.0 V battery. If you increase the voltage between the capacitor what would happen to the following quantities. Capacitance - Reason: Energy stored - Reason:

Explanation / Answer

here,

a)

the capacitance of capacitor when a dielectric is inserted between the plates , C= k * area * e0 /d

area - area of each plate

d - sepration between plates

k - dielectric constant

b)

area = 1 * 40 cm^2 = 40 * 10^-4 m^2

d = 0.2 m

k = 7

C= k * area * e0 /d

C = 7 * 40 * 10^-4 * 8.85 * 10^-12 /0.2 = 1.24 * 10^-13 F

the capacitance of capacitor when a dielectric is inserted between the plates is 1.24 * 10^-12 F

c)

V = 12 V

the magnitude of charge , Q = C * V

Q = 1.49 * 10^-11 C

d)

total energy stored , E = 0.5 * C * V^2

E = 0.5 * 1.24 * 10^-12 * 12^2

E = 8.93 * 10^-11 J

e)

C = k * area * e0 /d

so, when the voltage is increased

the capacitance remains the same

Energy stored , E = 0.5 * C * V^2

when voltage is increased

the energy stored in the capacitor also increases

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